2
$\begingroup$

This is a simple question, but I just don't understand how we determine the poles and zeros of a rational system function.

For example, for the LTI system described by this constant coefficient difference equation $$ y[n]-\frac{5}{2}y[n-1]+y[n-2]=x[n] $$ we can determine that

$$ H(z)=\frac{1}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})} $$

I understand why there are poles at $z=\frac{1}{2}$ and $z=2$, but I don't understand why there are two zeros at $z=0$. Even if I multiply $H(z)$ through by $z$, there would only be one zero, correct?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Just multiply numerator and denominator by $z^2$ to obtain

$$H(z)=\frac{z^2}{(z-\frac12)(z-2)}$$

from which you see that there's a double zero at $z=0$. Note that the number of zeros and poles is always equal if you include poles and zeros at infinity.

Improve this answer
$\endgroup$
3
  • $\begingroup$ If I multiplied the numerator and denominator by z, then wouldn't I end up with z/((z-1/2)(z-2))? Where does the z^2 come from? $\endgroup$
    – user50420
    Sep 30, 2018 at 17:16
  • $\begingroup$ @user50420: That was a typo, I meant "by $z^2$". $\endgroup$
    – Matt L.
    Sep 30, 2018 at 17:21
  • $\begingroup$ I see what you mean now. This was a trivial question, but for some reason I forgot how multiplication works. Thank for your help. $\endgroup$
    – user50420
    Sep 30, 2018 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.